Question

Number of millimoles in $1.0\text{ gram }$of water is :
(a) $1.0$
(b) $18$
(c) $55.56$
(d) $12$

Answer 1

Hint: We have to first find the moles of the water by using the given mass and by using the formula as $moles=\dfrac{given\text{ }mass}{molecular\text{ }mass}$ and after finding the moles multiply the obtained moles with $1000$ to get the millimoles. Now solve it.

Complete step by step solution:
First of all, let’s discuss the water molecule. Water is made up of two atoms: oxygen and hydrogen. It is polar in nature due to the presence of highly electronegative atoms i.e. the oxygen atom and the highly electropositive atom i.e. the hydrogen atom in it. It acts as a solvent for many organic compounds.
Now considering the statement;
First, we will have to find out the moles of water in $1.0\text{ gram }$ and then after that we will calculate its millimoles.
We can calculate the moles of water by using the formula as;
$moles=\dfrac{given\text{ }mass}{molecular\text{ }mass}$ -----------(1)
We know that;
Molecular mass of water is $=18$
Given mass of water=$1.0\text{ gram }$
Now, put these values in equation (1), we get;
$moles=\dfrac{1}{18}=0.05556\text{ moles}$------(2)
Now , to calculate the moles we will multiple equation (2) by $1000$, we get;
$\begin{align}
& moles=0.055\text{ }\times \text{1000 millimoles} \\
&\implies \text{ 55}\text{.56 millimoles} \\
\end{align}$
Thus, Number of millimoles in $1.0\text{ gram }$of water is $\text{=55}\text{.56 millimoles}$

Hence, option (c) is correct.

Note: According to the Avogadro’s law, one mole of an atom occupies $22.4\text{ L}$ of the volume and Avogadro number of particles i.e. $6.023\times 1{{0}^{23}}$ at standard conditions of temperature and pressure i.e. 1 atmosphere pressure and $273K$ temperature.