Molar mass of electron is nearly : `(N_(A) = 6 xx 10^(23))`

A compound contains 7 carbon atoms, 2 oxygen atoms and 1.0 xx 10^(-23) gmof other elements. The molecular mass of compound is : (N_(A) = 6 xx 10^(23))

Each molecular of a compound contains 5 carbon atoms, 8 hydrogen atoms and 4 xx 10^(-23) gm of other element. The molecular mass of compounds is (N_(A) = 6 xx 10^(23))

Each molecular of a compound contains 5 carbon atoms, 8 hydrogen atoms and 4 xx 10^(-23) gm of other element. The molecular mass of compounds is (N_(A) = 6 xx 10^(23))

An element with density 2.8 g cm^(-3) forms a fcc unit cell with edge length 4 xx 10^(-8) cm. Calculate the molar mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1) )

An element with density 2.8 cm^(3) forms a f.c.c unit cell with edge length 4xx10^(-8) cm. calculate the molar mass of the element. Given: (N_(A)=6.022xx10^(23)) mol^(-1)

An element with density 2.8 cm^(3) forms a f.c.c unit cell with edge length 4xx10^(-8) cm. calculate the molar mass of the element. Given: (N_(A)=6.022xx10^(23)) mol^(-1)

The number of electron in 3.1 mg NO_(3)^(-) is ( N_(A) = 6 xx 10^(23) )