If the r.m.s speed of chlorine molecule is $$490$$ m s$$^{-1}$$ at $$27°$$C, the r.m.s speed of argon molecules at the same temperature will be (Atomic mass of argon $$= 39.9$$ u, molecular mass of chlorine $$= 70.9$$ u)

The r.m.s. speed of a gas molecule is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

At the same temperature, the ratio of r.m.s. speeds of two gases is:

$$\frac{v_{Ar}}{v_{Cl_2}} = \sqrt{\frac{M_{Cl_2}}{M_{Ar}}}$$

Given: $$v_{Cl_2} = 490$$ m/s, $$M_{Cl_2} = 70.9$$ u, $$M_{Ar} = 39.9$$ u.

$$\frac{v_{Ar}}{490} = \sqrt{\frac{70.9}{39.9}}$$

Compute the ratio:

$$\frac{70.9}{39.9} = 1.7770$$

$$\sqrt{1.7770} = 1.3330$$

Therefore:

$$v_{Ar} = 490 \times 1.3330 = 653.2 \text{ m/s}$$

This is closest to $$651.7$$ m/s. Let me verify with more precision:

$$\frac{70.9}{39.9} = 1.77694...$$

$$\sqrt{1.77694} = 1.33302...$$

$$490 \times 1.33302 = 653.18$$

The closest option is $$651.7$$ m/s (Option B).

The answer is Option B: $$651.7$$ m/s$$^{-1}$$.