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A student takes a 60.0 mL volume from a stock solution of 0.532 mol L-1 magnesium chloride.How many moles of chloride ions, Cl−,  are present in the taken volume, 60.0 mL?Enter your answer with three significant figures. Do not include units.

Question

A student takes a 60.0 mL volume from a stock solution of 0.532 mol L-1 magnesium chloride.How many moles of chloride ions, Cl−,  are present in the taken volume, 60.0 mL?Enter your answer with three significant figures. Do not include units.

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Solution

First, we need to calculate the number of moles of magnesium chloride (MgCl2) in the 60.0 mL solution.

The molarity (mol/L) is given as 0.532 mol/L.

We know that molarity = moles/volume.

So, moles = molarity * volume.

But first, we need to convert the volume from mL to L because the molarity is given in mol/L.

1 L = 1000 mL.

So, 60.0 mL = 60.0/1000 = 0.060 L.

Now, we can calculate the moles of MgCl2.

moles = 0.532 mol/L * 0.060 L = 0.03192 mol.

Each unit of MgCl2 contains 2 units of Cl-.

So, the moles of Cl- = 2 * moles of MgCl2 = 2 * 0.03192 mol = 0.06384 mol.

Rounding to three significant figures, the moles of Cl- = 0.0638 mol.

This problem has been solved

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