G Salt

Question: What is the solubility (in g salt/100 g of H2O) of the salt? A 2.31 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C.

Answer: 2.31 : 9.10 = x : 100 x = 25.4 g/ 100 g H2O

Question: What is the solubility (in g salt/100 g of H2O) of the salt? A 4.35 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. What is the solubility (in g salt/100 g of H2O) of the salt? ___________ g

Answer: 4.35 : 9.10 = x : 100 x =47.8 g/100 g

Question: A sample consists of 55.0 g of salt and 35.0 g of sand. What is the purity of the salt? A sample consists of 55.0 g of salt and 35.0 g of sand. What is the purity of the salt? Please show how to arrive at the answer. Thanks!

Answer: The mixture is 61.1% salt 100 x 55.0 /(55.0+35.0) = 61.1

Question: What is the solubility of the salt (in g salt/ 100g of water) of the salt...? ...if 3.20g of salt dissolves in 9.10g of water to give a saturated solution at 25 degrees Celsius. I am not sure if this is right, but would I have to use molality to solve this question? Thank you!!

Answer: No, molality is not not needed here, just set up a proportion: 3.20g salt / 9.10g water = x g salt / 100g water Solving for x, x = 35.2 g salt. So the solubility is 35.2 g salt / 100g water.

Question: 15 g of salt to a beaker and add enough water to make 200ml of solutionbut realize its too concentrate,h? I add 15 g of salt to a beaker and add enough water to make 200ml of solution but realize its too concentrate,I should of used 8g of salt how much water must I add to obtain my desired concentration

Answer: If your solution should contain 8 g of salt in 200 ml of solution, then 1 g of salt would require 200/8 ml of solution and 15 g of salt requires (200/8) x 15 ml or 375 ml of solution. So you need to top up with water to the 375 ml level.

Question: How many g of salt are there in 32 mL of a 76%w/v salt solution? how many g of salt are there in 32 mL of a 76%w/v salt solution?

Answer: 1 g/mL = 100% w/v so 0.76 × 32 = 24.32 g salt ♣♦

Question: An impure sample of table salt that weighed 0.4652 g, the rest of the question is on the details? 5. An impure sample of table salt that weighed 0.4652 g, when dissolved in water and treated with excess AgNO3, formed 1.044 g of AgCl. What is the percentage of NaCl in the impure sample?

Answer: Write out the balanced equation for the reaction of salt with Ag nitrate. You know how much AgCl you got, so you can back calculate how much NaCl you actually started with. Calculate the % of the actual to the starting wt of 0.4652 g and you get the answer.

Question: Convert a salt concentration of 25 g/L in a solution of density 1005 kg/m3 to a mass fraction? heres a question i've got: Convert a salt concentration of 25 g/L in a solution of density 1005 kg/m3 to a mass fraction. hi, can u plz explain what equation you're using?

Answer: Mass fraction = (25g/1L) * (1m^3/1005kg) * (1kg/1000g) * (1000L/1m^3) =0.02488

Question: You have 125 g of a certain seasoning and are told that it contains 44.0 g of salt. What is the percentage of? salt by mass in this seasoning?

Answer: Percentage by mass is = mass of salt/ total mass of seasoning x 100% try working this out yourself. The answer should be 35.2%

Question: 3.51 g of salt dissolves in 9.10 g of H2O to give a saturated solution at 25°C. What is the solubility? What is the solubility (in g salt/100 g of H2O) of the salt? Please explain how to do this and give an answer. thanks :)

Answer: Just set up a proportion: 3.51 g salt / 9.10 g H2O = x g salt / 100 g H2O Solving, x = 38.6. The solubility is 38.6 g / 100 g H2O.

Question: a 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at? 25 degree celcius . what is the solubility 9in g salt/100 gof H2O) of the salt

Answer: 9.10 g of Water ( H2O ) can dissolve 3.20 g of Salt 100 g of H2O will dissolve . . . . 3.20 x 100 / 9.10 g of Salt = 35.1648 g Solubility = 35.2 g / 100 g of H2O

Question: What is the percentage strength of solution of table salt if 50 ml of water contains 10 g of salt?

Answer: Percentge strength? Do you mean % mass? Do you mean concentration? Well: water weights 18 g/mol and commom salt (sodium chloride) weights (23 + 35.5), then 58.5 g/mol. If you mean mass: Supposing normal conditions when water has a density of 1 g/cm3, then 50 ml water would weight 50g. In this case the solution would have 50g water and 10g salt so the % of salt would be: total weight = 50g water + 10 g salt = 60 g solution --> 10g salt would be 16.67 % If you mean concentration: 10 g salt would be 10/58.5 mol = 0.171 mol if dissolved in 50 ml ( = 0.05 litres) water, concentration will be --> 0.171/0.05 (mol/l or M) = 3.42 M 3.42 M (molar) = 3.42 mol per litre Please, next time be more specific about your doubts

Question: Hi, i want to set up a marine tank (nano) with clown fish but have no idea where to start, e.g salt level,PH? also what is ro water and when doing water changes how do you make sure it is safe to add

Answer: have a look at this site for setting up the tank and what is needed http://www.fishlore.com/SaltwaterAquariu… and this one will give ya a selection of fish and if they will mix with clowns.http://www.liveaquaria.com/ good luck:)

Question: if your STARTING a 55 g. salt water fish tank, where you spend big and where do you skimp?

Answer: The truth is that you don't need to spend big anywhere. This is a very marketable hobby and most sites really gear you in to spending way to much money. If you are just starting out for the first time, take it slow, read a lot, and check your local craigs list to see if you can jump on any setups from people who are getting out of the hobby. I built my own sump for about 30 dollars, my own protein skimmer for about ten dollars My lights I picked up on craigs list for about 150 dollars. You defiantly don't want to spend 3 thousand dollars on everything and not know how to maintain it or lose your animals. This guy sells rock for your tank and if you contact them they would be willing to help you set up all the necessities weather they help you build them or they locate what you need for you. They are very friendly so don't be afraid to contact them and ask questions about things other than the products they have listed. They helped me setup my tank and it has been up for 2 years now. I spent the most money on my live rock.http://www.kavatica.com/products.html and an earlier answerr seems to think that saltwater hobbying is only for the financially stable and that is a buch of tartar sauce, this hobby is for everyone and many people find out that is is more time involving than financial so go for it and don't let anyone tell you you can't afford it. Don't skimp on the live rock as that is your filter. You dont need all of it at once but you will need some.

Question: What is the percentage strength of a solution of table salt if 50 ml of water contains 10 g of salt? thank you so much

Answer: I would say 20%, but I'm not all too sure.

Question: What is the concentration of a solution prepared by mixing 50 g of salt with 200 mL of water? its for science

Answer: 50 g of salt with 200 mL of water? can be listed as 50 grams / 250 g total = a 20% salt solution by mass =========== if it is table salt, using its molar mass: 50 grams @ 58.44 g/mole = 0.855 moles 0.855 moles mixed with / 0.200 kilograms of water = 4.3 molal solution ====================== if you would l;ike molarity or some other concentration, email me

Question: A 3.42-g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C.? What is the solubility (in g salt/100 g of H2O) of the salt?

Answer: This is a simple ratio question. Our example gives us that 3.42 grams of salt saturate a 9.10 gram water sample. The ratio of salt to water is 3.42 / 9.10 = 0.376 g salt / g H2O. Now, the above ratio needs to be converted to account for 100 grams of water. So we multiply our answer by 100: (0.376) * (100) = 37.6 g salt / 100 g H2O

Question: how many teaspoons do you need to use for 400 g of salt? i need to make a 20% salt with water thing. this is the ingredient thing: 400 g of salt (NaCl) + 360 ml water

Answer: If you need the salt solution for an experiment, then don't even think of using teaspoons. Here's how to do it: 1.Weigh a beaker or any container. 2.Add 400 g to the mass of the container. 3.Place the beaker in a balance (platform, single pan, etc. ) 4.Put salt into the beaker until you reach the number you've calculated from 2. 5. Use graduated cylinder for the water. If you really need to use teaspoons, then... convert the amount of salt from grams to mL using the density of salt. mL= 400g/ 2.16. Then use your teaspoon. (but It's not going to be very accurate) A teaspoon is approximately 5mL.

Question: How many grams of table salt would have to dissolved in 145.0 g of water to lower the freezing point by 1.50 d? How many grams of table salt would have to dissolved in 145.0 g of water to lower the freezing point by 1.50 degrees Celsius ? kf = 1.86-kg/mol for water.

Answer: delta T = m x kf x i i = Vant'Hoff factor : for NaCl I = 2 1.50 = m x 1.86 x 2 m = 0.403 = moles NaCl / 0.145 Kg moles NaCl = 0.0585 mass NaCl = 0.0585 x 58.44 g/mol=3.42 g

Question: How many grams of table salt would have to dissolved in 110.0 g of water to lower the freezing point by 2.10? How many grams of table salt would have to dissolved in 110.0 g of water to lower the freezing point by 2.10 degrees celsius? kf = 1.86 K-kg/mol for water.

Answer: For these types of problems you need to know the equation for freezing point depression. It is delta T = imk where i is the van't Hoff constant (for table salt, NaCl, it will be = 2), m is molality (moles/kg solvent) and K is = 1.86.