Question

How many grams of magnesium cyanide are needed to make 275 mL of a 0.075
M solution?​

Answer 1

Explanation:

molarity=n / volume in L

0.075=n/275×1000

n=275/1000×0.075=20.625/1000 moles

molar mass of Mg(CN)2=76.3398g/mol

no of moles=mass / molar mass

20.625=m/76.3398

m=20.625×76.3398/1000=1.57450grams

Answer 2

Answer:

The mass of the magnesium cyanide measured is $1.52g$.

Explanation:

Given,

The volume of the solution, V = $275ml = 0.275L$

The molarity of the solution, M = $0.075M$

The mass of the magnesium cyanide, m =?

As we know,

• Molarity is defined as the number of moles of the solute per litre volume of the solution.
• Molarity = $\frac{n}{v}$

As shown here,

• n = The number of moles of solute
• V = The given volume of the solution

From the molarity equation, we have to find out the number of moles of solute.

Therefore,

• Molarity = $\frac{n}{v}$
• $0.075 = \frac{n}{0.275}$
• $n = 0.020 mol$

As we know,

• Number of moles = $\frac{Mass }{Molar mass}$

Here, the molar mass of the magnesium cyanide, $Mg(CN)_{2} = 76.34 g/mol$

After putting the values in the equation, we get:

• $0.020 = \frac{Mass}{76.34}$
• $Mass =1.52g$

Hence, the mass of the magnesium cyanide, m = $1.52 g$.