To dilute a solution from 0.600 M to 0.100 M, the final volume must be

Question

A. 1/6 the original volume.
B. 0.7 L.
C. More information is needed to answer this question.
D. 6 times the original volume.
E. 60 L.

Answer 1

Just pick a number like 100 mL
mL x M = mL x M
100 x 0.6 = mL x 0.1
Solve for mL and I get
100*0.6/0.1 = 600
600 mL is what times 100?

Answer 2

d

Answer 3

The concentration of Pb2+ in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of sodium carbonate, forming the insoluble lead (II) carbonate (267 g/mol) according to the balanced equation given below. The solid lead (II) carbonate is dried, and its mass is measured to be 0.1443 g. What was the concentration of Pb2+ in the original wastewater sample?

Pb2+(aq) + Na2CO3(aq) → PbCO3(s) + 2Na+(aq)

Group of answer choices