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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?
Wiki User
∙ 14y ago
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
Wiki User
∙ 14y ago
To find the molarity of the resulting solution, we first need to convert the volume of water to liters: 300.0 mL = 0.3 L. Next, we calculate the total volume of the resulting solution: 0.3 L (water) + 0.4 L (initial solution) = 0.7 L. Finally, we calculate the moles of Na2CrO4 in the initial solution (0.4 L * 0.400 M = 0.16 moles), and divide by the total volume to find the molarity: 0.16 moles / 0.7 L = 0.23 M.
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