Questions & Answers

Question

Answers

(A)- 41.35

(B)- 42.35

(C)- 41.65

(D)- 41.75

Answer
Verified

Let us see what we have been given.

Number of molecules of X in the mixture = $1.65\times {{10}^{21}}$

Number of molecules of Y in the mixture = $1.85\times {{10}^{21}}$

Total mass of the mixture = 0.688g

Molecular mass of Y= 187

Molecular mass of Y is equal to the mean mass of one molecule of Y which is 187 (in atomic mass unit).

We know that one mole of a substance contains $6.022\times {{10}^{23}}$ molecules.

Therefore, one mole of Y contains $6.022\times {{10}^{23}}$ molecules and the mass of one mole of Y (in grams) is equal to the average or mean mass of one molecule of Y (in amu), i.e. 187.

Hence, we can write that

Mass of $6.022\times {{10}^{23}}$ molecules of Y = 187g

Then, we can say that the mass of $1.85\times {{10}^{21}}$ molecules of Y = $\dfrac{187g}{6.022\times {{10}^{23}}}\times 1.85\times {{10}^{21}}$

Simplifying the above equation gives the mass of $1.85\times {{10}^{21}}$ molecules of Y = 0.574g.

Since the total mass of the mixture is 0.688g, such that

Mass of $1.65\times {{10}^{21}}$ molecules of X + mass of $1.85\times {{10}^{21}}$ molecules of Y = 0.688g

Then, mass of $1.65\times {{10}^{21}}$ molecules of X = 0.688g - mass of $1.85\times {{10}^{21}}$ molecules of Y

Now, we have calculated the mass of $1.85\times {{10}^{21}}$ molecules of Y to be 0.574g.

Therefore, on subtracting 0.574g from 0.688g, we get

Mass of $1.65\times {{10}^{21}}$ molecules of X is = 0.114g.

Thus, mass of $6.022\times {{10}^{23}}$ molecules of X = $\dfrac{0.114g}{1.65\times {{10}^{21}}}\times 6.022\times {{10}^{23}}$

Simplifying the above equation and calculating we obtain, mass of $6.022\times {{10}^{23}}$ molecules of X = 41.60g.

Since, the mass of one mole of X (in grams), i.e. 41.60g is equal to the average or mean mass of one molecule of X (in amu). Therefore, the molecular mass of X is 41.60g.

The only option close to 41.60 is (C).