Answer: C (3 salts). Every salt here is made by neutralising the dilute nitric acid, so all the salts produced are nitrates. Lead carbonate reacts with nitric acid to give lead(II) nitrate, \(\text{PbCO}_3 + 2\text{HNO}_3 \to \text{Pb(NO}_3)_2 + \text{H}_2\text{O} + \text{CO}_2\). Copper carbonate reacts similarly to give copper(II) nitrate. Aqueous sodium hydroxide is neutralised by nitric acid to give sodium nitrate, \(\text{NaOH} + \text{HNO}_3 \to \text{NaNO}_3 + \text{H}_2\text{O}\).
That makes three different salts: lead(II) nitrate, copper(II) nitrate and sodium nitrate. The two carbonates do not react usefully with sodium hydroxide to give a new salt, so no further salts are possible.
So three salts can be prepared, giving answer C.
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