If 50. mL of 1.0 M NaOH is diluted with distilled water to a volume of 2.0 L, the concentration of the resulting solution is A0.025 M B 0.050 M C 0.10 M D 0.50 M E 1.0 M M

The concentration of the resulting solution if 50. mL of 1.0 M NaOH is diluted with distilled water to a volume of 2.0 L is 0.025M

HOW TO CALCULATE FINAL CONCENTRATION:

C1V1 = C2V2

Where;

According to this question;

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The concentration of the resulting solution is 0.025 M

Explanation:

According to the dilution law,

where,

= molarity of concentrated solution = 1.0 M

= volume of concentrated solution = 50 ml

= molarity of diluted solution = ?

= volume of diluted solution = 2.0 L = 2000 ml (1L=1000ml)

Putting in the values we get:

Therefore, concentration of the resulting solution is 0.025 M

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o