Calulate the molesules mass of a substance whose 7.0% by mass solution in water freezes at `-0.93^(@)C`. the cryosctopic constant of water is `1.86^(@)C kg mol^(-1)` :

Calculate the molecular weight of a substance whose 7.0% (by weight) solution in water freezes at -0.89^(@)C . (k_(f) "for" H_(2)O=1.86K"molality"^(-1) ).

6 grams of a substance 'A' dissolved in 100 g of water freezes at - 0.93^(@)C . The molecular mass of 'A' is ( K_(f)=1.86 Km^(-1) ).

When 36g of a non-volatile, non-electrolytic solute having the empirical formula CH_(2)O is dissolved in 1.2 kg of water, the solution freezes at -0.93^(@)C . The molecular formula of the solute is ( K_(f) of water = 1.86 K kg mol^(-1) )

When 36.0 g of a solute having the empirical formula CH_(2)O is dissovled in 1.20 kg of water, the solution freezes at -0.93^(@)C . What is the moleculer formula of the solute ? ( K_(f) = 1.86^(@)C kg mol^(-1) )

When 36.0 g of a solute having the empirical formula CH_(2)O is dissovled in 1.20 kg of water, the solution freezes at -0.93^(@)C . What is the moleculer formula of the solute ? ( K_(f) = 1.86^(@)C kg mol^(-1) )

When 36.0 g of a solute having the empirical formula CH_(2)O is dissovled in 1.20 kg of water, the solution freezes at -0.93^(@)C . What is the moleculer formula of the solute ? ( K_(f) = 1.86^(@)C kg mol^(-1) )

What is osmotic pressure of the aq. Solution of the given solute at 27^(@)C if depression in freezing point is 0.93^(@)C ? Molal depression constant of water is 1.86K mol^(-1) kg ? Assume molality as molarity.

The freezing point of an aqueous solutions is 272 .93 K . Calculate the molality of the solution if molal depression constant for water is 1.86 Kg mol^(-1)