how many ml of 10.0 m cacl2 do we need to prepare 3.00 l of a 0.500 m cacl2 solution

Question

Question

in progress 0

Chemistry Hồng Cúc 4 years 2021-08-17T03:18:16+00:00 2021-08-17T03:18:16+00:00 1 Answers 30 views 0

Answers ( )

Name*

E-Mail*

Website

Featured image

Browse

Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

Answer*

danthu · Answer 1 · 2021-08-17T03:20:01+00:00

Answer:

150 mL

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 10 M

Volume of diluted solution (V₂) = 3 L

Molarity of diluted solution (M₂) = 0.5 M

Volume of stock solution needed (V₁) =?

The volume of stock solution needed can be obtained as follow:

M₁V₁ = M₂V₂

10 × V₁ = 0.5 × 3

10 × V₁ = 1.5

Divide both side by 10

V₁ = 1.5 / 10

V₁ = 0.15 L

Finally, we shall convert 0.15 L to mL. This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.15 L = 0.15 L × 1000 mL / 1 L

0.15 L = 150 mL

Therefore, the volume of the stock solution needed is 150 mL