Answer to Question #188379 in General Chemistry for Olivia Odigie

Question

How many kilograms of strontium nitrate would I need to make 52.1 liters of a 0.750 M solution? 

Answer 1

Moles of Sr(NO₃)₂₌(molarity of Sr(NO₃)₂ in m)(vol in L)

=(0.750M)(52.1l)

M= mol.l^-1

=(0.750mol.l^-1(52.1l)

=39.075 mol

Molar mass of Sr(NO₃)₂₌ 211.63g/mol

Mass of Sr(NO₃)₂=(39.075 mol)(211.63g/mol)

=8.27*10³g

=(8.27*10³/1000) kg

=8.27Kg