$2 K _2 CrO _4+2 HNO _3(\text { dil }) \rightarrow K _2 Cr _2 O _7 2 KNO _3+ H _2 O$

Therefore $; CrO _4^{2-}+2 HNO _3($ dil $) \rightarrow Cr _2 O _7^{2-}+2 NO _3^{-}+ H _2 O$

Dilute $HNO _3$ is an oxidising agent. In this reaction oxidation no. of $Cr$ changes from $+3$ to $+6$. So, dilute $HNO _3$ oxidises $K _2 CrO _4$ to $K _2 Cr _2 O _7$