Mike mowed 3 1/4 lawns and allen mowed 2 5/6 lawns. how many more lawns did allen mow?

Answer: 30 grams

Step-by-step explanation: (50 mL x x%) = (1000 mL x 0.3%)

x = 6%

6% x 500 mL = 30 grams

Final answer:

The calculation for the amount of sodium chloride (NaCl) in a solution depends on the desired concentration. For instance, to prepare 50 ml of a 1M solution, one would need 2.922 g of NaCl. For a 500 ml stock solution for the same, the measurement would be ten times this, or 29.22 g.

Explanation:

The question pertains to preparing a solution of sodium chloride N(aCl). The given information indicates that we have 5.30 mol NaCl L solution. To provide an accurate answer, it's necessary to know the targeted concentration for the 50 ml solution. However, lacking this data, we can consider an example where we want to prepare a 1M solution.

In such a case, using the concept of molarity (mol/L), we'd first establish how many moles of NaCl are needed. For a 1M solution, we'd need 1 mol of NaCl per liter of solution. Therefore, in order to prepare 50 ml (or 0.05 L), we require 0.05 mol of NaCl.

The given information also states that 1 mol of NaCl weighs 58.44 g. Therefore, we would need (0.05 mol) * (58.44 g/mol) = 2.922 g of NaCl for 50 ml of a 1M solution. Therefore, if preparing 500 ml of a stock solution, we would require 10 times this amount, or 29.22 g of NaCl.

This is a general guide as the specific quantity can vary based on the desired concentration of the 50 ml solution. For other concentrations, we would use the same method, simply adjusting the moles of NaCl needed as appropriate.

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