Calculate molarity given density and mass percentage
Solution
By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$:
$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$
The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$:
$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$
The amount of substance $n$ can be found using mass of solute $m$ and molar mass of solute $M$.
Knowing mass fraction $\omega,$ one can switch to the mass of solution $m_\mathrm{tot}$ already used in \eqref{eqn:2}:
$$n = \frac{m}{M} = \frac{\omega m_\mathrm{tot}}{M}\label{eqn:3}\tag{3}$$
By plugging both \eqref{eqn:2} and \eqref{eqn:3} into \eqref{eqn:1}, one receives the final expression:
$$
\require{cancel}
\begin{align}
c &amp;= \frac{\omega \cancel{m_\mathrm{tot}}\rho}{M\cancel{m_\mathrm{tot}}} \\
  &amp;= \frac{\omega\rho}{M} \\
  &amp;= \frac{0.27\times\pu{1.198 g cm-3}}{\pu{98.08 g mol-1}} \\
  &amp;= \pu{3.30E-3 mol cm-3}~\text{or}~\pu{3.30 mol L-1}\tag{4}
\end{align}
$$
Notes

There is little to no point of converting anything prior to calculations as if the dimensional analysis is done right, you'll get the answer anyways.
Density is an intensive property: you cannot use mass fraction to determine the density of solute alone.
Regarding &quot;no of moles&quot;: first, don't use non-standardized abbreviations like &quot;no&quot;, and second, it's amount of substance, not &quot;number of moles&quot;.
&quot;Number of moles&quot; is an appropriate term in a dialogue with your neighbor about the population of talpids in your garden, but not in science.
Don't forget to define each physical quantity properly and always try to solve the problem algebraically first, and plug in the values at the end when you have the ready-to-use equation for the quantity you were asked to find.
Keep the track of significant figures (slang: &quot;sigfigs&quot;), e.g. you are given the density with four sigfigs, so it would make sense to also use molar mass with four sigfigs as well.
Yet due to two sigfigs of mass fraction, the accuracy of the final answer is reduced.