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300.0 Ml of water is added to .40 L of a .400 M Na2CrO4 solution what is the molarity of the resulting solution?
300.0 ml of water is added to .40 L of a .400 M Na2CrO4 solution what is the molarity of the resulting solution?
Na2CrO4 = 2 Cr +Cr + 4 O's
Molar mass = (2*23 + 52 + (4*16) = 162
A .400 M N Na2CrO4 solution has .400 moles of Na2CrO4 in a liter of water.
.400 moles of Na2CrO4 = 0.400 * 162 = 64.8 grams of Na2CrO4 in a liter of water.
Since you only have .40 L, you have 64.8 grams/liter * 0.4L = 25.92 grams of Na2CrO4 in 0.4 liter of solution.
When you add 300.0 ml of water, you have total of 700 ml of solution.
You still have 25.92 grams of Na2CrO4, but now you have 700 ml of solution.
Molarity = moles of solute per liter of solution.
Moles of solute = grams of solute ÷ Molar mass of solute
Moles of solute = 25.92 ÷ 162 = 0.16 moles of Na2CrO4.
Molarity = 0.16 moles of Na2CrO4 ÷ 0.700 L of solution.
Molarity = 0.23 M
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