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$6.84 \mathrm{~g}$ of sucrose is dissolved in $200 \mathrm{~g}$ of water. The molality of the solution is
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$0.1 \mathrm{M}$
Molar mass of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)=342$
Molality $=\frac{\frac{\text { Meight of sucrose }}{\text { Molar mass of sucrose }}}{\text { Mass of solvent }(g)} \times 1000$
$=\frac{6.84 \times 1000}{342 \times 200}=0.1 \mathrm{M}$
Molality $=\frac{\frac{\text { Meight of sucrose }}{\text { Molar mass of sucrose }}}{\text { Mass of solvent }(g)} \times 1000$
$=\frac{6.84 \times 1000}{342 \times 200}=0.1 \mathrm{M}$
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