A $20 \cdot g$ $20g$ mass of $C a C l_{2}$ $CaCl_2$ was dissolved in $700 \cdot g$ $700g$ of water. What is the molality of the solution with respect to $calcium chloride$ $"calcium chloride"$ ?

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A $20 \cdot g$ $20g$ mass of $C a C l_{2}$ $CaCl_2$ was dissolved in $700 \cdot g$ $700g$ of water. What is the molality of the solution with respect to $calcium chloride$ $"calcium chloride"$ ?

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Answer 1 · 2017-06-15T15:03:43

$0.25$ $0.25$ $m o l a l$ $molal$

Explanation:

Chemical formula of Calcium chloride= $C a C l_{2}$ $CaCl_2$
Molar mass of $C a C l_{2}$ $CaCl_2$ = $110.98$ $110.98$ $g m o l^{- 1}$ $gmol^-1$
Mass of Calcium chloride (given) = $20 g$ $20g$

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

No. of moles= $\frac{20g}{110.98g/mol}$ $("20g")/"110.98g/mol"$ = $0.180$ $0.180$ moles
$g$ $g$ of solvent = $700 g$ $700 g$
Converting $g$ $g$ $\to$ $->$ $k g$ $kg$
$k g$ $kg$ of solvent = $0.7 k g$ $0.7kg$
Water is solvent while $C a C l_{2}$ $CaCl_2$ is solute
Now,

$M o l a l i t y$ $Molality$ = $m$ $m$ = $\frac{moles of solute}{kg of solvent}$ $("moles of solute")/"kg of solvent"$
$M o l a l i t y$ $Molality$ = $\frac{0.180 moles}{0.7kg}$ $("0.180 moles")/"0.7kg"$
$M o l a l i t y$ $Molality$ = $0.25$ $0.25$ $m o l a l$ $molal$

Answer 2 · 2017-06-15T15:18:33

$Concentration \equiv 0.257 \cdot molal$ $"Concentration"-=0.257*"molal"$ .........

Explanation:

$Molality$ $"Molality"$ $\equiv$ $-=$ $\frac{Moles of solute}{Kilograms of solvent}$ $"Moles of solute"/"Kilograms of solvent"$

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

$molality = \frac{\frac{20 \cdot g}{110.98 \cdot g \cdot m o l^{- 1}}}{700 \cdot g \times 10^{- 3} \cdot k g \cdot g^{- 1}} = 0.257 \cdot m o l \cdot k g$ $"molality"=((20*g)/(110.98*g*mol^-1))/(700*gxx10^-3*kg*g^-1)=0.257*mol*kg$ .

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

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A $20 \cdot g$ $20g$ mass of $C a C l_{2}$ $CaCl_2$ was dissolved in $700 \cdot g$ $700g$ of water. What is the molality of the solution with respect to $calcium chloride$ $"calcium chloride"$ ?

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A 20*g20⋅g20*g mass of CaCl_2CaCl2CaCl_2 was dissolved in 700*g700⋅g700*g of water. What is the molality of the solution with respect to "calcium chloride"calcium chloride"calcium chloride"?

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A $20 \cdot g$ $20g$ mass of $C a C l_{2}$ $CaCl_2$ was dissolved in $700 \cdot g$ $700g$ of water. What is the molality of the solution with respect to $calcium chloride$ $"calcium chloride"$ ?