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What mass of Na₂SO4 is needed to make 2.44 L of 4.368 M solution? (Na = 23 g; S = 32 g; 0 = 16 g)a. 254.2b. 1513c. 0.07506d. 0.01261

User Cagmz
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1 Answer

3 votes

Answer:

To solve this question, we need to use the following equation:

M = n/V

where:

M = molarity/concentration

n = number of moles

V = volume

Then, we use m = n*MM

So:

M = 4.368 mol/L

V = 2.44 L

n = ?

4.368 = n/2.44

n = 4.386*2.44

n = 10.65792 moles

MM of Na2SO4 = (2*23) + (1*32) + (4*16) = 142 g/mol

m = 10.65792 *

m = 1,513 g

Answer: b. 1513

User Cybergoof
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