Yes, I was talking about what is described in your resources. You can do this:
// define a floating-point literal in hex and print it in decimal
float x = 0x1p-8; // x = 1.0/256
printf("x = %g\n", x); // prints 0.00390625
// define a floating point literal in decimal and print it in various ways
float y = 0.3; // non-representable, rounded to closest float
printf("y = %g\n", y); // 0.3 (the %g format does some heuristics)
printf("y = %.10f\n", y); // 0.3000000119
printf("y = %.20f\n", y); // 0.30000001192092895508
printf("y = %a\n", f); // 0x1.333334p-2
Your question is ambiguous for two different reasons. First, this value is not representable as a floating-point number, so there's no way that you can even store it in a float. Second, once you have a float variable, you can print it in many different ways. So, the answer to your question is, irremediably, "it depends what you mean by exact value".
If you print your variable with the %a format, then YES, the exact value is preserved and there is no loss of information. The problem is that the literal that you wrote cannot be represented exactly. But this is hardly a fault of the floats. Ints have exactly the same problem:
int x = 2.5; // x gets the value 2
int y = 7/3; // same thing