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3 years ago

What is the molarity of a solution that contains 1 mol CaCl2 dissolved in 2 L of solution?

Chemistry

1 answer:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

molarity= 0.5M

Explanation:

n=CV

n= 1mol, C= ?, V= 2L

1= C×2

C= 0.5M

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3 years ago

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3 years ago

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3 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

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3 years ago