Question

To go forth with a reaction in a laboratory 250.0 ml of a solution is needed from an 0.03690 M in AgNo3 acqueose solution. How many grams of AgNo3 with a 87.90 pureness are needed to prepare the solution. Molar mass of AgNo3= 169.87g
Options:
A. 1.567 g
B. 6.268 g
C. 1.783 g
D. 7.132 g
E. 1.377 g

Answer 1

V = 250.0 mL = 0.25 L

C = 0.0369 mol/L

MM = 169.87 g/mol

$n = V \times C \\ n = 0.25 \times 0.0369 = 0.009225 \;mol \\ m = n \times M \\ m = 0.009225 \times 169.87 = 1.567 \;g$

Proportion:

1.567 – 87.90 %

x – 100 %

$x = \frac{1.567 \times 100}{87.90} = 1.783 \;g$

Answer: C. 1.783 g