Question

A student wished to prepare 150.0 mL of a 0.490 M AgNO3 solution. What volume (in mL) of 1.95 M AgNO3 would she need to take?

Answer 1

Using formula:$N_1V_1=N_2V_2$

$0.490\times150=1.95\times V_2$

Hence volume of $AgNO_3$ required$=V_2=37.69ml$