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Q: What is the molarity of a solution that contains 125 g of nacl in 400 l of solution?

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125g NaCl x 1 mol NaCl / 58.44 NaCl= 2.13894 mol / 4.00 L = 0.535 M

Molarity is moles per litre, however we have 4 litres. So we have to divide 125 by 4 first. 125/4 is 31.25 grams. The molecular weight of NaCl is 58.443g/mol so the overall molarity is 0.53471 molar.

The gram formula mass (for an ionically bonded compound such as NaCl, the counterpart to a gram mole for covalently bonded compounds) is 58.44. Therefore, 125 g of NaCl constitutes 125/58.44 or about 2.1389 gram "moles". Since molarity is defined as the ratio of moles of solute dissolved to total volume of solution, the given solution is 0.535, to the justified number of significant digits.

First off you have to divide by 274 and multiply by 1000 to get the grams in a litre. This gives you 456.2g per litre. Then you divide by the molecular mass. 456.2/58.443 which is 7.8 molar.

sucrose is c12h22o11 -- molar mass - 12x12 + 22x1 + 11x16 = 342 g/mole (125 / 342) / 3.5 = 0.1044 M

chota

The molarity of a solution made by dissolving 23,4 g of sodium sulphate in enough water to make up a 125 mL solution is 1,318.

The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution

0.0125

The percentage of a 1 to 125 solution is simply 1/125 or 0.008 or 0.8%.

Molarity = moles of solute/volume of solution 0.500 M KOH = moles KOH/125 ml 62.5 millimoles, or to answer the question precisely, 0.0625 moles KOH

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