200 cc of $$ x\times 10^{-3} M$$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic mediun.
Here $$x$$ =___________

According to the law of equivalence, 

The milliequivalents of the oxidant  = ($$K_2Cr_2O_7$$, $$n=6$$) 

must equal those of the reductant (Mohr's salt, $$n=1$$). 

Using the formula:  $$(M_1 \times n_1 \times V_1) = (M_2 \times n_2 \times V_2)$$, 

we get $$(x \times 10^{-3} \times 6) \times 200 = (0.6 \times 1) \times 750$$. 

This simplifies to $$1.2x = 450$$, 

which yields $$x = 375$$.