# A 20*g mass of CaCl_2 was dissolved in 700*g of water. What is the molality of the solution with respect to "calcium chloride"?

Jun 15, 2017

$0.25$ $m o l a l$

#### Explanation:

Chemical formula of Calcium chloride= $C a C {l}_{2}$
Molar mass of $C a C {l}_{2}$ =$110.98$ $g m o {l}^{-} 1$
Mass of Calcium chloride (given) = $20 g$

$M o l a l i t y$ = $m$ = ("moles of solute")/"kg of solvent"
Now, to calculate no. of moles:
No. of moles = ("mass")/"molar mass"

No. of moles= ("20g")/"110.98g/mol" = $0.180$ moles
$g$ of solvent = $700 g$
Converting $g$ $\to$ $k g$
$k g$ of solvent = $0.7 k g$
Water is solvent while $C a C {l}_{2}$ is solute
Now,

$M o l a l i t y$ = $m$ = ("moles of solute")/"kg of solvent"
$M o l a l i t y$ = ("0.180 moles")/"0.7kg"
$M o l a l i t y$ = $0.25$ $m o l a l$

Jun 15, 2017

$\text{Concentration"-=0.257*"molal}$.........

#### Explanation:

$\text{Molality}$ $\equiv$ $\text{Moles of solute"/"Kilograms of solvent}$

And thus..............................................

$\text{molality} = \frac{\frac{20 \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1}}{700 \cdot g \times {10}^{-} 3 \cdot k g \cdot {g}^{-} 1} = 0.257 \cdot m o l \cdot k g$.

Note that at (relatively!) low concentrations, the calculated $\text{molality}$ would be almost the same as solution $\text{molarity}$. $\text{Molality}$ is used because the expression is fairly independent of temperature. What is the $\text{molal concentration}$ with respect to $\text{chloride ion}$?