Experts...the mass of MgCl2 sholud be dissolved in 750 gm of water in order to prepare a 1.05 m solution is...?

Given: 
mass of solvent, that is , water = 750 
molality = 1.05 m
Thus, using the formula :
molality = $\frac{number of moles of solute \times 1000}{mass of solvent }$
1.05 = $\frac{number of moles of MgC{l}_{2_{ }}\times 1000}{750}$
number of moles of MgCl_​2 = 0.7875 moles
Now, number of moles = mass / molar mass 
0.7875 =  mass /95 
mass of MgCl₂ = 74.81 g
Thus, 74.81 g of MgCl₂ is required to dissolve in 750 g of water to make 1.05 molal solution.