Density of 1M solution of a non-electrolyte `C_(6)H_(12)O_(6)` is 1.18 g/mL . If `K_(f) (H_(2)O)` is `1.86^(@) mol^(-1)` kg , solution freezes at :

Density of 1M solution of a non-electrolyte C_(6)H_(12)O_(6) is 1.18g//mL . If K_(f)(H_(2)O) is 1.86^(@)kg , solution freezes at :

Density of 1M solution of C_(6)H_(12)O_(6) is 1.18 g//cm^(3) . If K_(f)(H_(2)O) is 1.86 K m^(-1) , the solution freezes at

Mole fraction of a non-electrolyte in aqueous solution is 0.07 . If K_(f)(H_(2)O) is 1.86^(@) mol^(-1) kg , depression in f.p ., DeltaT_(f) is :

What is the freezing point of a solution contains 10.0g of glucose C_(6)H_(12)O_(6) , in 100g of H_(2)O ? K_(f)=1.86^(@)C//m

What is the freezing point of a solution contains 10.0g of glucose C_(6)H_(12)O_(6) , in 100g of H_(2)O ? K_(f)=1.86^(@)C//m

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molarity of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molality of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molarity of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

The freezing point of 0.05 molal solution of non- electrolyte in water is _____. (K_(f) = 1.86 ^(@)Ckg mol ^(-1))

What would be the freezing point of aqueous solution containing 17 g of C_(2)H_(5)OH in 100 g of water (K_(f) H_(2)O = 1.86 K mol^(-1)kg) :