Let the pressure inside the balloon be $p_i$ and the outside pressure be $p_0$, then excess pressure is $p_i-p_0=\frac{2 S}{r}$.

where, $S=$ Surface tension

$r=$ radius of balloon

Considering the air to be an ideal gas $p_i V=n_i R T_i$ where, $V$ is the volume of the air inside the balloon, $n_i$ is the number of moles inside and $T_i$ is the temperature inside, and $p_0 V=n_0 R T_0$ where $V$ is the volume of the air displaced and $n_0$ is the number of moles displaced and $T_0$ is the temperature outside.

So, $n_i=\frac{p_i V}{R T_i}=\frac{M_i}{M_A}$

where, $M_i$ is the mass of air inside and $M_A$ is the molar mass of air and

$$n_0=\frac{p_0 V}{R T_0}=\frac{M_0}{M_A}$$

where, $M_0$ is the mass of air outside that has been displaced. If $w$ is the load it can raise, then $w+M_i g=M_o g$

$$\Rightarrow \quad w=M_o g-M_i g$$

As in atmosphere $21 \% \mathrm{O}_2$ and $79 \% \mathrm{~N}_2$-is present

$\therefore$ Molar mass of air

$$M_i=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}$$

$\therefore$ Weight raised by the balloon

$$w=\left(M_o-M_i\right) \mathrm{g}$$

$$\begin{aligned} \Rightarrow \quad \mathrm{w} & =\frac{M_A V}{R}\left(\frac{p_0}{T_0}-\frac{p_i}{T_i}\right) g \\ & =\frac{0.02884 \times \frac{4}{3} \pi \times 8^3 \times 9.8}{8.314} \quad\left(\frac{1.013 \times 10^5}{293}-\frac{1.013 \times 10^5}{333}-\frac{2 \times 5}{8 \times 313}\right) \\ & =\frac{0.02884 \frac{4}{3} \pi \times 8^3}{8.314} \times 1.013 \times 10^5\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8 \\ & =3044.2 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} \therefore \quad \text { Mass lifted by the balloon } & =\frac{w}{g}=\frac{3044.2}{10} \approx 304.42 \mathrm{~kg} . \\ & \approx 305 \mathrm{~kg} . \end{aligned}$$