What mass of Zn(NO3)2 is needed to make 0.53 L of 0.77 M solution.

Answer 1

Molar mass of Zn(NO3)2 = 189.36 g/mol

0.77 M means you want 0.77 moles Zn(NO3)2 in 1 liter of solution.

Since you don't need to make 1 liter, but only need to make 0.53 L, we find the moles needed for that.

0.77 mol/L x 0.53 L = 0.4081 moles needed

Now we simply convert this to grams using the molar mass...

0.4081 moles x 189.36 g/mol = 77.28 g = 77 g (rounding to 2 significant figures)