# What mass of solute is required to prepare a 0.875*"molal" solution from 534*mL of water solvent?

Aug 12, 2016

Approx. $27 \cdot g$. I assume you mean a $\text{molal concentration}$.

#### Explanation:

$\text{Molality"="moles of solute"/"kg of solvent}$

Thus $0.875 \cdot m o l \cdot k {g}^{-} 1$ $=$ $\frac{\text{number of moles of NaCl}}{0.534 \cdot k g}$

And $\text{number of moles of NaCl}$ $=$ $0.875 \cdot m o l \cdot k {g}^{-} 1 \times 0.534 \cdot k g$ $=$ $0.467 \cdot m o l$

$\text{Grams of NaCl} = 0.467 \cdot m o l \times 58.44 \cdot g \cdot m o {l}^{-} 1$. The $\text{molar concentration}$ would be to all intents and purposes identical.