516 mL of a 3.82 M sodium sulfate ( $Na_2SO_4$ ) solution is diluted with 0.875 L of water. What is the new concentration in molarity?

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Answer 1 · 2016-04-26T18:27:26

Approx. $1.4*mol*L^-1$

$"Molarity "= " Moles of solute"/"Volume of solution (L)"$

$=$ $(516xx10^-3*cancelLxx3.82*mol*cancel(L^-1))/(0.516*L+0.875*L)$

$=$ $(1.97*mol)/(1.391*L)$

$~=$ $1.4*mol*L^-1$

516 mL of a 3.82 M sodium sulfate (Na_2SO_4Na_2SO_4) solution is diluted with 0.875 L of water. What is the new concentration in molarity?