Answer
Verified
394.2k+ views
Hint: Calculate the mass of molecules of Y present in the mixture. Eliminating the mass of molecules of Y from the total mass of the mixture will give us the mass of all the molecules of X in the mixture.
Complete answer:
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).
Hence, the correct option is (C).
Note: Molecular mass is generally expressed in atomic mass unit (amu) whereas the unit for molar mass is gram/mol. We are likely to make calculation mistakes, so carefully solve the question step by step to avoid any errors.
Complete answer:
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).
Hence, the correct option is (C).
Note: Molecular mass is generally expressed in atomic mass unit (amu) whereas the unit for molar mass is gram/mol. We are likely to make calculation mistakes, so carefully solve the question step by step to avoid any errors.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE