Answer:
D. 1.406 m
Explanation:
Hello,
At first we need a basis, that could be: [tex]m_{solution}=100g[/tex], secondly, based on the initial basis, to get the 13.5% by mass, it implies that the mass of the solute (calcium chloride) is 13.5g:
[tex]m_solute=m_{solution}*m/m=100g*0.135=13.5g_{solute}[/tex]
Thus, the mass of the solvent in kg is:
[tex]m_{solvent}=m_{solution}-m_{solute}=100g-13.5g=86.5g*\frac{1kg}{1000g} \\m_{solvent}=0.0865kg[/tex]
Now, the moles of the solute:
[tex]molCaCl_2=13.5g*\frac{1molClCl_2}{110.68g CaCl_2} =0.122mol CaCl_2[/tex]
Finally, the molality results:
[tex]m=\frac{mol_{solute}}{m_{solvent}}=\frac{0.122mol}{0.0865kg}\\\\m=1.41m[/tex]
Which pretty close to the D answer.
Best regards.
