Question

13.9 Molarity and chemical reactions in aqueous solution Volume of solution A Volume of solution B Molarity Molarity Particle
Percent by mass (g solute/g solution) Take 100 g solution Mass of solution Mass of solute Density Molar mass Volume of soluti
26) 1250 ml. of 400 M Calle A) 0.400 mole of Cats B) 1.00 mole of Cabra CaBr, is diluted to 1.60 L. C) 2.40 moles of Cabr D)
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Answer #1

26)

Dilution will not effect the number of mol.

Number of mol after dilution = number of mol before solution

= M(CaBr2)*V(CaBr2) in L initially

= 4.00 M * 0.250 L

= 1.00 mol

Answer: B

27)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 1.36 M

V1 = 850 mL

V2 = 4500 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (1.36*850)/4500

M2 = 0.257 M

Answer: E

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