The molecular weight of Na is 58.44 grams/mole. To prepare a 0.50 M solution: how many grams of NaCI would you need in 1,000 mL of water?

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The molecular weight of Na is 58.44 grams/mole. To prepare a 0.50 M solution: how many grams of NaCI would you need in 1,000 mL of water?

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Answer 1 · 2016-08-18T20:34:51

$Molarity$ $"Molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$ , so here we need $29.22 \cdot g$ $29.22*g$ of $N a C l$ $NaCl$

Explanation:

And thus, $\frac{29.22 \cdot g}{58.44 \cdot g \cdot m o l^{- 1}} \times \frac{1}{1 \cdot L}$ $(29.22*g)/(58.44*g*mol^-1)xx1/(1*L)$ $=$ $=$ $0.500 \cdot m o l \cdot L^{- 1}$ $0.500*mol*L^-1$ .

You might ask as to whether the volume of water would change (up or down upon dissolution of the salt. The volume change would be imperceptible.

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The molecular weight of Na is 58.44 grams/mole. To prepare a 0.50 M solution: how many grams of NaCI would you need in 1,000 mL of water?

1 Answer

Explanation:

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