The freezing point of a solution prepared from $1.25 \ g$ of a non-electrolyte and $20 \ g$ of water is $271.9 \ K$. If the molar depression constant is $1.86 \ K \ kg \ mol^{-1}$,then the molar mass of the solute will be:

Calculate the molar mass of a non-volatile solute when $1 \ g$ of it is dissolved in $100 \ g$ of solvent,which decreases its freezing point by $0.2 \ K$. Given: $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.

$3 \times 10^{-3} \ kg$ acetic acid is added into $500 \ cm^{3}$ water. If dissociation of acetic acid is $23\%$ then find out depression in freezing point? $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$ and density $= 0.997 \ g \ cm^{-3}$.

When an aqueous solution of glucose is frozen,which crystals will separate out first?

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

$1.00 \text{ g}$ of a non-electrolyte solute dissolved in $50 \text{ g}$ of benzene lowered the freezing point of benzene by $0.40 \text{ K}$. The freezing point depression constant of benzene is $5.12 \text{ K kg mol}^{-1}$. Find the molar mass of the solute.