Answer
$5.375\times10^{24}\;\rm molecule$
Work Step by Step
First, we need to find the number of molecules of the nitrogen gas which is given by the ideal gas law.
$$PV=nRT$$
Hence,
$$n=\dfrac{PV}{RT}\tag 1$$
Now we need to find the number of nitrogen molecules which is given by
$$N=nN_{\rm A}$$
where $N_{\rm A}$ is Avogadroe's number.
Plugging from (1) and then plugging the known;
$$N=\dfrac{PV}{RT}N_{\rm A}=\dfrac{(1.013\times 10^5)(1)}{(8.31)(20+273}(6.02\times 10^{23})$$
Hence,
$$N=2.50\times 10^{25}\;\rm molecule$$
Now we need to use (Figure 18.2) which shows the distribution of molecular speeds in a sample of nitrogen gas.
It shows that about 11% of the gas is at a speed of 700-800 m/s.
Also, it shows that about 7% of the gas is at a speed of 800-900 m/s.
And, about 3.5% of the gas is at a speed of 900-1000 m/s.
Hence, in our sample, the number of gas at a speed of 700-100 m/s is given by
$$N_1=(0.11+0.07+0.035)N=0.215(2.50\times 10^{25})$$
$$N_1=\color{red}{\bf 5.375\times10^{24}}\;\rm molecule$$
