Freezing point

[parwana]

How many grams of urea (MW = 60.056 g/mol) would have to be dissolved in 115.0 grams of water to lower the freezing point by 1.60 degrees celsius?

Equation for freezing point
Tf= Kf x Cm

Kf= 1.858
Cm= molality of solute in kg

please help, iam getting the same answer

[chem_tr]

Here, T_f is 1.60, and from there you can calculate the molality:

T_f=K_f \times C_m and C_m=\frac {T_f}{K_f}

C_m=\frac {1.60}{1.858}=0.861

Since your molality is 0.861, it means that in 1 kg of solution you'll have to dissolve moles of this amount of compound.

[parwana]

i still dont kn ow how to calculate the grams required, please help

[chem_tr]

What did you find as mole amount from the last equation I gave? You'll just multiply this value with molar mass of urea to learn how many grams should be in the solution.