# Question #24349

##### 1 Answer

#### Explanation:

Here's my *interpretation* of the problem

The specific volume of a cylindrical virus particle is equal to#6.02 * 10^(-2)"cm"^3"g"^(-1)# .The radius of the particle is equal to#7Å# and its height is equal to#10Å# .What is the molecular mass of the virus?

If this is indeed what your problem looks like, then your starting point here will be the **specific volume**,

The *specific volume* of a given substance is defined as the ratio between the **volume** of the substance and the **mass** it occupies. In your case, the virus is said to have a specific volume of

#v = 6.02 * 10^(-2)"cm"^3"g"^(-1)#

This means that **for every** gram of virus particles you get a volume of

Now, the virus is said to be shaped like a **cylinder**.

The volume of a cylinder is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)V = pi * r^2 * hcolor(white)(a/a)|)))#

In your case, the radius of the cylinder,

#r = 7Å = 7 * 10^(-10)"m"#

The height of the cylinder is said to be

#h = 10Å = 10 * 10^(-10)"m"#

Since the specific volume of the virus is expressed in *cubic centimeters* per gram, convert the radius and height of the cylinder to *centimeters*

#7 * 10^(-10) color(red)(cancel(color(black)("m"))) * (10^2"cm")/(1color(red)(cancel(color(black)("m")))) = 7 * 10^(-8)"cm"#

and

#10 * 10^(-10)color(red)(cancel(color(black)("m"))) * (10^2"cm")/(1color(red)(cancel(color(black)("m")))) = 10^(-7)"cm"#

The volume of one virus particle will this be

#V = pi * (7 * 10^(-8)"cm")^2 * 10^(-7)"cm"#

#V = 1.54 * 10^(-21)"cm"^3#

Use the specific volume of the virus to calculate the **mass** of one particle

#1.54 * 10^(-21) color(red)(cancel(color(black)("cm"^3))) * "1 g"/(6.02 * 10^(-2)color(red)(cancel(color(black)("cm"^3)))) = 2.56 * 10^(-20)"g"#

In order to find the **molecular mass** of the virus, you need to find the mass of **one mole** of virus particles.

As you know, **Avogadro's number** gives you the number of particles present in one mole

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"particles" color(white)(a/a)|))) -># Avogadro's number

Use the mass of **one particle** to find the mass of *one mole* of particles

#2.56 * 10^(-20)"g"/color(red)(cancel(color(black)("particle"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("particles"))))/"1 mole" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.5 * 10^(4)"g mol"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that the values given to your for the radius and height of the cylinder justify only one sig fig.