The density of gold is $19 \ g/cm^3$. If $1.9 \times 10^{-4} \ g$ of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius $10 \ nm$,then the number of gold particles per $mm^3$ of the sol will be

  • A
    $1.9 \times 10^{12}$
  • B
    $6.3 \times 10^{14}$
  • C
    $6.3 \times 10^{10}$
  • D
    $2.4 \times 10^6$

Explore More

Similar Questions

If the mean of a Poisson distribution is $\frac{1}{2}$,then the ratio of $P(X=3)$ to $P(X=2)$ is

In which of the following compounds does nitrogen exhibit the highest oxidation state?

The area (in square units) of the region bounded by the curves $x=y^2$ and $x=3-2y^2$ is

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

The gravitational force acting on a particle,due to a solid sphere of uniform density and radius $R$,at a distance of $3 R$ from the centre of the sphere is $F_1$. $A$ spherical hole of radius $(R / 2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. Ratio of $F_1$ and $F_2$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo